What is the solution for y^{''}+3y^'-4y=2 ?

The solution for y^{''}+3y^'-4y=2 is y=c_{1}e^t+c_{2}e^{-4t}-1/2

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y^{''}+3y^'-4y=2

y^{^{''}} + 3 y^{^{'}} - 4 y = 2

y = c_{1} e^{t} + c_{2} e^{- 4 t} - \frac{1}{2}

Solution steps

y^{''} (t) + 3 y^{'} (t) - 4 y = 2

Solve linear ODE:

y = c_{1} e^{t} + c_{2} e^{- 4 t} - \frac{1}{2}

y = c_{1} e^{t} + c_{2} e^{- 4 t} - \frac{1}{2}

y^{^{''}} + 4 y = - 5

y^{^{''}} + 2 y^{^{'}} + 4 y = 3 cos (2 t) + 4 sin (2 t)

y^{^{''}} + 4 y = e^{x} + x^{2} + 7 x

y^{^{'''}} - 4 y^{^{''}} + 5 y^{^{'}} - 2 y = 6 e^{x} + 10 cos (x)

y^{^{''}} + 3 y^{^{'}} + 2 y = sin (x)

What is the solution for y^{''}+3y^'-4y=2 ?
The solution for y^{''}+3y^'-4y=2 is y=c_{1}e^t+c_{2}e^{-4t}-1/2